Gulpjs wait for async function before continue

Question

I cut gulpfile.js into several files. My purpose is to share the data with various tasks. And I use windows system, gulp 5 and "type": "module" config.

The file structure is given below:

gulpfile.js
tasks/clean.js
tasks/deploy_data.js
data/path.js
data/other.js

gulpfile.js

import gulp from 'gulp';
import clean from './tasks/clean.js';
import deploy_data from './data/deploy_data.js';

global.data = {}; // share data with all tasks.

function deploy_at_start() {
  gulp.series(deploy_data);
}

deploy_at_start(); // I hope that deploy_data will be executed before any task I debug, because each task needs to use the contents of data.

console.log(data); // {} Blank, obviously deploy_data has not been completed yet, I want the code below to be executed only after it completes.

export {
  clean,
  // alot of other tasks
};

./tasks/clean.js

function clean(cb) {
  console.log(data.path.src); // I need to get some information from data, such as the path I want to delete.
  cb();
}

export default clean;

The point is this

data/path.js

const path = {
  src: './folder/*',
};

export default path;

tasks/deploy_data.js

import path from 'node:path';
import { pathToFileURL } from 'node:url';

async function deploy_data(cb) {
  const dir = path.join(process.cwd(), 'data');
  const files = fs
.readdirSync(dir, { withFileTypes: true }).filter((item) => item.isFile())
.map((item) => item.name); // I have a lot of data files, so this is the easier way to get all the data

  for (const file of files) {
    data[path.parse(file).name] = (await import(pathToFileURL(path.join(dir, file)).href)).default;
  }

  cb();
}

export default deploy_data;

When I execute gulp clean, I can't get the data.path information. I think it may be because deploy_data is async and something went wrong.

From the terminal, I can see that deploy_data completes its work after clean is completed, which is why data is empty.

I understand that I can use gulp.series to execute deploy_data before clean, but I have a huge amount of data. I think I can achieve the goal by executing deploy_data once at the beginning of gulpfile.js similar to the current method.

In the above case, is there a way for me to execute deploy_data before executing any gulp task (the task that has been exported) and wait for it to complete before executing other code?

Answer 1

Just add a top-level await:

await deploy_at_start();

and don't forget to return from deploy_at_start so await would wait gulp.series() to finish:

function deploy_at_start() {
  return gulp.series(deploy_data);
}